Chapter 4 Properties of the Integers

1

Discrete and Combinatorial Mathematics

R. P. Grimaldi, 5th edition, 2004

Chapter 4Properties of the Integers

2

Well-Ordering Principle (良序原理 )

Every nonempty subset A of + = {1,2,…} contains a smallest element.

Note: A can be finite or infinite.

Some sets are not well-ordered: : A=

, : A=(0,1)

It is the basis of a prove technique -- Mathematical Induction.

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Principal of Mathematical Induction

Let S(n) denote an open mathematical statement that involves one or more occurrences of the variable n +.

(a) If S(1) is true; and (b) If S(k) is true for some particular k +, then S(k+1) is true;then S(n) is true for all n +.

4

Proof of Theorem 4.1

Proof by contradiction:

Define the set F = { m | S(m) does not hold} +. If F is non-empty, then F must have a smallest element m (well-ordering of +), with S(m). Because we know that S(1), it must hold that m>1.

Because m is the smallest value, it must hold that S(m–1), which contradicts our proof for all k +: S(k)S(k+1).

Thus F has to be empty.

Therefore, S holds for all +. #

5

Principal of Strong Form Mathematical Induction

Let S(n) denote an open mathematical statement that involves one or more occurrences of the variable n +. Let n0, n1 + with n0 n1.

(a) If S(n0), S(n0+1), S(n0+2), …, S(n1-1), and

S(n1) are true; and

(b) If whenever S(n0), S(n0+1), …, S(k-1),and S(k)

are true for some particular k +, where k

n1, then the statement S(k+1) is also true;

Then S(n) is true for all n n0.

6

Mathematical Induction

If we want to prove a property P(n) for all n+, we can do that in the following inductive way:

Basis step:Prove it for n=1: does P(1) hold?

Inductive step:Assuming that P(1), P(2), …, P(k) holds for some k+, prove P(k+1).

By the structure of +, this proves it for all n:

P(1) holds, hence P(2), hence P(3), et cetera.

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Example of Induction

Show that n! 2n-1, for n +.

Proof.

Induction Basis. n = 1 1! 21-1 = 1.

Inductive step.

Assume that i! 2i-1, for i = 1, 2, …, k. (k+1)! = (k+1)k!. Then

(k+1)! = (k+1)k! (k+1) 2k-1

2 2k-1 = 2k.

Therefore, the assertion is true. #

8

More Example of Induction

For every n+ where n14, prove S(n) : n can be written as a sum of 3’s and/or 8’s.

Proof.

Induction Basis. n = 14 n = 3 + 3 + 8.

n = 15 n = 3 + 3 + 3 + 3 + 3.

n = 16 n = 8 + 8.

Inductive step.

Assume that S(14), S(15), …, S(k) are true

for some k+ where k16. k+1 = (k-2) + 3. S(k+1) is true.

Thus, S(n) is true for all n14. #

9

Recursive Definition

Fibonacci sequence: 0,1,1,2,3,5,8,13,21,… It is sequence number A000045 in The On-Line Encyclopedia of Integer

Sequences at http://www.research.att.com/~njas/sequences/index.html

Recursive base. F0 = 0, F1 = 1.

Recursive process. Fn = Fn-1 + Fn-2, n 2.

Explicit formula:

For large n, it grows like Fn ≈ 0.447214 1.61803n.

This (1+√5)/2 ≈ 1.61803 is the Golden Ratio.

nn

n 2

51

5

1

2

51

5

1F

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Fibonacci in NatureShape of shells:

Golden ratio:

Perfect shape : total height / navel height = 1.618

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Property of Fibonacci

Observation.

Conjecture.

11110 2221

20 FF

212110 22222

21

20 FFF

3262110 222223

22

21

20 FFFF

531532110 2222224

23

22

21

20 FFFFF

8540532110 22222225

24

23

22

21

20 FFFFFF

.10

2

nn

n

ii FFF

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Property of Fibonacci

Proof.

Induction Basis.

Inductive step.

Assume that

for some k+ where k1.

#

21222

120 11110 FFFF

10

2

kk

k

ii FFF

211

21

0

21

0

2

kkkk

k

ii

k

ii FFFFFF

.)( 2111 kkkkk FFFFF

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Other Sequences

Harmonic sequence:

Lucas sequence:

)1

2

1

1

1(,),

3

1

2

1

1

1(),

2

1

1

1(),

1

1(

k

.,1),1

1(

,1

1

1

nnn

HH

H

nn

,29,18,11,7,4,3,1,2

.,2,

,1,2

21

10

nnLLL

LL

nnn

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More Sequences

Binomial sequence:

Eulerian sequence:

0,0),(;,0),(;1)0,0(

0),1,(),(),1(

rrnCnrrnCC

rnrnCrnCrnC

.0,0;,0;1

10,)1()(

,,0,0

,11,1,

kamkaa

mkakakma

kmkm

kmkmkm

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Number SystemBinary digits: 0 and 1, called bits.

Review of decimal system: Example: 45,238 is equal to

8 ones 8 x 1 = 8

3 tens 3 x 10 = 30

2 hundreds 2 x 100 = 200

5 thousands 5 x 1000 = 5000

4 ten thousands 4 x 10000 = 40000

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From Binary to Decimal

The number 1101011 is equivalent to 1 one 1 x20 = 1 1 two 1x21 = 2 0 four 0x22 = 0 1 eight 1x23 = 8 0 sixteen 0x24 = 0 1 thirty-two 1x25 = 32 1 sixty-four 1x26 = 64

107 in decimal base

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From Decimal to Binary

The number 7310 is equivalent to 73 = 2 x 36 + remainder 1 36 = 2 x 18 + remainder 0 18 = 2 x 9 + remainder 0 9 = 2 x 4 + remainder 1 4 = 2 x 2 + remainder 0 2 = 2 x 1 + remainder 0

7310 = 10010012

(write the remainders in reverse order preceded by the quotient 1)

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Adding Binary Numbers

Example: add 1001012 + 1100112

1 1 1 carry ones

1001012

+1100112

10110002

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Hexadecimal Number System

Decimal system0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 1 2 3 4 5 6 7 8 9 A B C D E F

Hexadecimal system

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Hexadecimal to Decimal

The hexadecimal number 3A0B16 is

11 x 160 = 11

0 x 161 = 0

10 x 162 = 2560

3 x 163 = 12288

1485910

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Two’s Complement

One’s complement: Replace each 0(1) in the binary representation by 1(0).

100101

011010

Two’s complement: Add 1 to one’s complement.

011010

+ 1

011011

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Subtraction

33 - 15

33 = 001000012, 15 = 000011112

-15 = 111100012

001000012

+111100012

(1)000100102 = 18

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More Subtraction

15 - 33

33 = 001000012, 15 = 000011112

-33 = 110111112

000011112

+110111112

111011102

Two’s complement

000100102 = 18

15 – 33 = -18.

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Divisibility of Integers

Let x,yZ and y0, x is a multiple of y if and only if there exist and integer mZ such that x = ym.

We also say y divides x or y is a divisor of x.

Notation: y|x.

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Division Properties

For all a,b,c,x,y,z Z:

a) 1|a and a|0

b) [(a|b) and (b|a)] a = ±b

c) [(a|b) and (b|c)] a|c

d) a|b a|bx for all x

e) x = y + z, a|x and a|y a|z, a|y and a|z a|x

f) [(a|b) and (a|c)] a|(bx+cy) bx+cy is a linear combination of b and c.

g) a|ci, cjZ, for 1 ≤ i ≤ n

a|(c1x1+c2x2 + … + cnxn) for all xjZ.

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Division Algorithm

Let a,bZ and b>0, there exist unique q,rZ with 0≤r<b, such that a = qb + r.

We call a the dividend, b the divisor, q the quotient, and r the remainder.

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Primes

An integer p>1 is prime if and only if its two positive divisors are 1 and p.

An integer n>1 that is not prime is composite.

The first primes are: 2,3,5,7,11,…

Euclid: there are infinitely many primes.

Prime factorization is unique for each n{2,3,4,…}.

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Property of Composite Integers

Lemma 4.1 If n Z+ is composite, then there is a prime p such that p|n.

Proof. Let S be the set of all composite integers that have no prime divisors.

Assume S is not empty.

Let m be the least member of S.

m is composite m1,m2 [m = m1 m2].

p [p|m1], where p is prime.

p| m1 m2 p|m. #

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Infinitely Many Primes

Theorem 4.4 There is an infinite number of primes.

Proof. Assume there are finite primes p1, p2, …,pk.

Consider N = p1 p2 … pk + 1.

N must be a composite.

By Lemma 4.1, pi [pi | N].

However piN for 1 i k. #

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Property of Composite Integers

If n Z+ is composite, then there is a prime p such that p|n and .

Proof.

n is composite n1,n2 Z+ [n = n1 n2].

Assume and .

n1 n2 > n.

Without loss of generality, we assume

By Lemma 4.1, a prime p [p | n1].

p|n and . #

np

nn 1 nn 2

.1 nn

np

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Common Divisor

For a,b, a positive integer c is said to be a common divisor of a and b if c|a and c|b.

The greatest common divisor c, denoted by gcd(a,b), is the common divisor of a and b such that for any common divisor d of a and b satisfies d|c.

Examples: gcd(121,33) = 11; gcd(6,35)=1, gcd(8,16)=8.

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Uniqueness of GCD

For all a,bZ+, there exists a unique cZ+ that is the greatest common divisor of a and b.

Proof.Given a,bZ+, let S = {as+bt | s,t, as+bt>0}.

We claim that the smallest element cS is gcd(a,b).

cS x,yZ [c = ax + by].

Assume ca. a = qc + r, qN,rZ+, 1 r < c.

r = a – qc = a – q(ax+by) = (1-qx)a – (qy)b.

rS contradicts to that c is the smallest in S.

Thus, c|a. Similarly, c|b.

If dZ and d|a and d|b, then d|c. (Rule (f)) #

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Properties of GCD

For all a,b+, gcd(a,b) is the smallest positive integer we can write as a linear combination of a and b.

Integers a and b are called relatively prime when gcd(a,b) = 1. That is, x,y [ax + by = 1].

Examples: gcd(42,70) = 14 x,y [42x + 70y = 14] x,y [3x + 5y = 1] x = 2-5k, y = -1+3k. (Infinite solutions.)

Theorem 4.8 If a,b,c+, the Diophantine equation ax+by=c has an integer solution iff gcd(a,b)|c.

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Euclidean Algorithm

Let a,b+. gcd(a,b) is calculated by setting r0 = a, r1 = b, and applying the division algorithm n times as follows:r0 = q1r1 + r2, 0<r2<r1

r1 = q2r2 + r3, 0<r3<r2

…

ri = qi+1ri+1 + ri+2, 0<ri+2<ri+1

…

rn-2 = qn-1rn-1 + rn, 0<rn<rn-1

rn–1 = qnrn. gcd(a,b)=rn

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Least Common Multiple

For a,b,c+, c is a common multiple of a and b if a|c and b|c. Furthermore, c is the least common multiple of a,b, denotd by lcm(a,b), is the smallest of all common multiples of a and b.The lcm(a,b) always exists and it is unique.

For any common multiple d of a and b, lcm(a,b)|d.

For all a,b+, ab=gcd(a,b)lcm(a,b).

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Irrrational Number

is irrational.Proof. (Aristotle,384-322 B.C.)Assume = a / b, for some a,b+ and gcd(a,b)=1. 2 = a2 / b2

2b2 = a2

2|a2 2|aLet a = 2c. 2b2 = a2 = 4c2

b2 = 2c2

2|b gcd(a,b) 2. Thus, is irrational. #

2

2

2

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Prime Factoring

Prime factoring of 980220 is 22 3 5 17 312.

The number of positive divisors of 980220 is(2+1)(1+1)(1+1)(1+1)(2+1) = 72.

Let m = 22 3 5 17 312, n = 23 7 52 31.gcd(m,n) = 22 5 31lcm(m,n) = 23 3 52 7 17 312.

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Uniqueness of Prime Factoring

Every integer n > 1 can be written as a product of primes uniquely. (The Fundamental Theorem of Arithmetic)Proof. Existence.Assume m is the smallest integer not expressible as a product of primes.

m is not a prime m1,m2+ [m = m1 m2]

m1,m2 < m m1 and m2 can be written as products of primes m can be written as a product of primes.

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Uniqueness of Prime Factoring Uniquness.Basis step: n = 2 can be uniquely written as a product of primes.

Inductive step: Assuming that 2, 3, …, n-1 can be uniquely written as products of primes.Suppose ,where

p1 | n p1 |

p1 | qj for some 1 j r

p1 = q1 by contradiction. (p1 < pe = q1 < qj = p1)

By induction, n can be uniquely written as a product of primes. #

rk tr

ttsk

ss qqqpppn 2121

2121 rk qqqppp 2121 ,

rtr

tt qqq 21

21

nqqqppppnn rk tr

ttsk

ss 2121

21

121

111 /

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Chapter 4 Properties of the Integers

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